Question

A particular water sample contains 2.0 X 10^-2 M Cu⁺ and 8.0 x 10^-3M Ca²⁺. You are asked to separate these two cations through selective precipitation. You are given a bottle of Nal. What cation will precipitate first? Calculate the percent of precipitated cation left in solution when you have precipitated as much of it you can without beginning the precipitation of the other cation.

Answer 1

Cu+ and Ca2+ both react with NaI to give CuI and CaI2   respectively.

Cu+ +I-↔CuI, ksp=1*10^-12

Ca2+ + I-↔CaI2 ,ksp=7.1*10^-7

As solubility product,ksp of CaI2>>>ksp CuI ,so CaI2 is more soluble,hence CuI will be precipitated first.

For calculating percent of precipitated cation left in the solution

Ksp=[Cu+][I-]=7.1*10^-7

If the solubility of Cu+=S

Then [Cu+]=S=[I-] in solution

Or, S*S=Ksp=7.1*10^-7

S^2=7.1*10^-7=0.71*10^-8

S=0.843 mol/L

So CuI left in the solution=0.843 mol/L

Initial conc=2.0*10^-2 mol/L

percent of precipitated cation left in the solution=0.843/2.0*10^-2*100=42.15%